This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A001451 #48 Oct 16 2024 09:23:18
%S A001451 1,20,1260,100100,8817900,823727520,79919739900,7962100660800,
%T A001451 808906548235500,83426304143982800,8707404737345073760,
%U A001451 917663774856743842200,97491279924241456098300,10427604345391237790688000,1121786259855036145008408000
%N A001451 a(n) = (5*n)!/((3*n)!*n!*n!).
%C A001451 From _Peter Bala_, Jul 15 2016: (Start)
%C A001451 This sequence occurs as the right-hand side of the binomial sum identity Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = (-1)^m*a(m) for n = 2*m. The sum vanishes for n odd. Cf. A273628 and A273629.
%C A001451 Note the similar results:
%C A001451 Sum_{k = 0..n} (-1)^k*binomial(n,k)* binomial(2*n + k,n)*binomial(3*n + k,n) = (-1)^n*(3*n)!/n!^3 = (-1)^n*A006480(n);
%C A001451 Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n)^2 = A002894(n);
%C A001451 Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n - k,n) = binomial(2*n,n) = A000984(n);
%C A001451 Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n + k,n)*binomial(3*n - k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(2*n - k,n)*binomial(3*n + k,n) = (-1)^n*binomial(2*n,n) = (-1)^n*A000984(n). (End)
%C A001451 Choose three noncollinear step vectors to satisfy the zero sum, 3*v_1 + v_2 + v_3 = 0. Then a(n) is the number of loop plane walks of length 5*n which depart from and return to the origin. Equivalently, a(n) counts distinct permutations of a (5*n)-digit integer with digits 1,2,3 of multiplicity 3*n,n,n respectively. - _Bradley Klee_, Aug 12 2018
%H A001451 Vincenzo Librandi, <a href="/A001451/b001451.txt">Table of n, a(n) for n = 0..200</a>
%F A001451 a(n) = binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n](F(x)^(20*n)), where F(x) = 1 + x + 12*x^2 + 390*x^3 + 16984*x^4 + 867042*x^5 + 48848541*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. - _Peter Bala_, Jul 14 2016
%F A001451 a(n) ~ 3^(-3*n-1/2)*5^(5*n+1/2)/(2*Pi*n). - _Ilya Gutkovskiy_, Jul 13 2016
%F A001451 G.f.: G(x) = 4F3(1/5,2/5,3/5,4/5;1/3,2/3,1;(5^5/3^3)*x). Let G^(n)(x) = d^n/dx^n G(x), and c = {120, 15000*x-6, 45000*x^2-114*x, 25000*x^3-135*x^2, 3125*x^4-27*x^3}, then Sum_{n=0..4} c_n*G^(n)(x) = 0. - _Bradley Klee_, Aug 12 2018
%F A001451 From _Peter Bala_, Mar 20 2022: (Start)
%F A001451 Right-hand side of the following identities valid for n >= 1:
%F A001451 Sum_{k = 0..3*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (5*n)!/((3*n)!*n!^2);
%F A001451 Sum_{k = 0..n} 4*n*(4*n+k-1)!/(k!*n!*(3*n)!) = (5*n)!/((3*n)!*n!^2). Cf. A000897 and A113424. (End)
%F A001451 a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(4*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - _Peter Bala_, Oct 12 2024
%e A001451 G.f. = 1 + 20*x + 1260*x^2 + 100100*x^3 + 8817900*x^4 + 823727520*x^5 + ... - _Michael Somos_, Aug 12 2018
%p A001451 f := n->(5*n)!/((3*n)!*n!*n!);
%t A001451 Table[(5*n)!/((3*n)!*n!*n!), {n, 0, 20}] (* _Vincenzo Librandi_, Sep 04 2012 *)
%o A001451 (Magma) [Factorial(5*n)/(Factorial(3*n)*Factorial(n)*Factorial(n)): n in [0..30]]; // _Vincenzo Librandi_, May 22 2011
%o A001451 (GAP) List([0..15],n->Factorial(5*n)/(Factorial(3*n)*Fact0rial(n)*Factorial(n))); # _Muniru A Asiru_, Aug 12 2018
%Y A001451 Cf. A000897, A002894, A002897, A006480, A008977, A186420, A188662, A273628, A273629.
%K A001451 nonn,easy,walk
%O A001451 0,2
%A A001451 _N. J. A. Sloane_